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Question Number 162366 by cortano last updated on 29/Dec/21

 Given that the solution set of the    quadratic inequality ax^2 +bx+c >0   is (2,3). Then the solution set    of the inequality cx^2 +bx+a <0    will be

$$\:{Given}\:{that}\:{the}\:{solution}\:{set}\:{of}\:{the}\: \\ $$ $$\:{quadratic}\:{inequality}\:{ax}^{\mathrm{2}} +{bx}+{c}\:>\mathrm{0} \\ $$ $$\:{is}\:\left(\mathrm{2},\mathrm{3}\right).\:{Then}\:{the}\:{solution}\:{set}\: \\ $$ $$\:{of}\:{the}\:{inequality}\:{cx}^{\mathrm{2}} +{bx}+{a}\:<\mathrm{0}\: \\ $$ $$\:{will}\:{be}\: \\ $$

Answered by mr W last updated on 29/Dec/21

a<0  x^2 +(b/a)x+(c/a)=(x−2)(x−3)  (b/a)=−(2+3)=−5 ⇒b=−5a  (c/a)=2×3=6 ⇒c=6a  cx^2 +bx+a=a(6x^2 −5x+1)<0  since a<0,  ⇒6x^2 −5x+1>0  (2x−1)(3x−1)>0  ⇒x<(1/3) or x>(1/2)  i.e. x∈(−∞,(1/3)) ∧ ((1/2),+∞)

$${a}<\mathrm{0} \\ $$ $${x}^{\mathrm{2}} +\frac{{b}}{{a}}{x}+\frac{{c}}{{a}}=\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right) \\ $$ $$\frac{{b}}{{a}}=−\left(\mathrm{2}+\mathrm{3}\right)=−\mathrm{5}\:\Rightarrow{b}=−\mathrm{5}{a} \\ $$ $$\frac{{c}}{{a}}=\mathrm{2}×\mathrm{3}=\mathrm{6}\:\Rightarrow{c}=\mathrm{6}{a} \\ $$ $${cx}^{\mathrm{2}} +{bx}+{a}={a}\left(\mathrm{6}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{1}\right)<\mathrm{0} \\ $$ $${since}\:{a}<\mathrm{0}, \\ $$ $$\Rightarrow\mathrm{6}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{1}>\mathrm{0} \\ $$ $$\left(\mathrm{2}{x}−\mathrm{1}\right)\left(\mathrm{3}{x}−\mathrm{1}\right)>\mathrm{0} \\ $$ $$\Rightarrow{x}<\frac{\mathrm{1}}{\mathrm{3}}\:{or}\:{x}>\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $${i}.{e}.\:{x}\in\left(−\infty,\frac{\mathrm{1}}{\mathrm{3}}\right)\:\wedge\:\left(\frac{\mathrm{1}}{\mathrm{2}},+\infty\right) \\ $$

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